11 ms·

> Fahrenheit (since this is a non-linear transform) huh?

by bhousel 12y ago

icegreentea 12y ago
There's a constant offset between F and C, so it's not linear in the actual engineering/math (or whatever) sense.
- drdeca 12y ago
  It's not proportional, but the equation for going from one to another is that of a line. I would think this would be called linear, but that mean something more specific?
  - icegreentea 12y ago
    In engineering, linearity refers to having two properties: a) f(x+y) = f(x) + f(y) where x, y, are two values, and f is a linear transform/function b) af(x) = f(ax) where a is constant In this case, the transform is the conversion between C and F. Since there is +/32 in there depending on direction, neither of these properties hold. For example, yards to meters is a linear transform. 3 yards converted to meters + 2 yards converted to meters = 5 yards converted to meters. This is not true of temperature. Adding 32F to 32F gives 64F, which is not the same as adding 0C to 0C.
    - drdeca 12y ago
      Thanks! I wasn't aware of that(the) definition.
  - ppereira 12y ago
    In geometry, this would be an affine transformation.
fluffyllemon 12y ago
For a function to be a linear transform it needs two properties[1]: 1. F(c1 + c2) = F(c1) + F(c2) 2. F(k*c1) = k*F(c1) Converting from Celsius->Fahrenheit (or vice versa) is does not have these properties. F(c) = 9/5 * c + 32 F(c1 + c2) = 9/5 * (c1 + c2) + 32 F(c1) + F(c2) = (9/5 * c1 + 32) + (9/5 * c2 + 32) = 9/5 * (c1 + c2) + 64 F(k*c1) = 9/5 * (k*c1) + 32 k*F(c1) = k * (9/5 * c1 + 32) = 9/5 * (k*c1) + k*32 Admittedly, I'm not entirely sure why being a non-linear transform gets the OP so riled up. [1]http://mathworld.wolfram.com/LinearTransformation.html http://mathworld.wolfram.com/LinearTransformation.html
- raverbashing 12y ago
  " Admittedly, I'm not entirely sure why being a non-linear transform gets the OP so riled up." Because it's harder to calculate mentally One can approximate 2lb to 1kg roughly, 1mi ~ 1.6km, but for Fahrenheit this is harder One solution is to establish a set point ( 70F = 21.1C, approximate to 20C ) then work in differences, since 1.8deltaF = 1deltaC so, 2dF ~ 1dC (or 10dF ~ 5dC) Hence, 90F is approximately 30C (exact value, 32.2C)

> Fahrenheit (since this is a non-linear transform) huh?

by bhousel 12y ago

> Fahrenheit (since this is a non-linear transform) huh?

icegreentea 12y ago

There's a constant offset between F and C, so it's not linear in the actual engineering/math (or whatever) sense.

drdeca 12y ago
It's not proportional, but the equation for going from one to another is that of a line. I would think this would be called linear, but that mean something more specific?
- icegreentea 12y ago
  In engineering, linearity refers to having two properties: a) f(x+y) = f(x) + f(y) where x, y, are two values, and f is a linear transform/function b) af(x) = f(ax) where a is constant In this case, the transform is the conversion between C and F. Since there is +/32 in there depending on direction, neither of these properties hold. For example, yards to meters is a linear transform. 3 yards converted to meters + 2 yards converted to meters = 5 yards converted to meters. This is not true of temperature. Adding 32F to 32F gives 64F, which is not the same as adding 0C to 0C.
  - drdeca 12y ago
    Thanks! I wasn't aware of that(the) definition.
- ppereira 12y ago
  In geometry, this would be an affine transformation.

fluffyllemon 12y ago

For a function to be a linear transform it needs two properties[1]: 1. F(c1 + c2) = F(c1) + F(c2) 2. F(k*c1) = k*F(c1) Converting from Celsius->Fahrenheit (or vice versa) is does not have these properties. F(c) = 9/5 * c + 32 F(c1 + c2) = 9/5 * (c1 + c2) + 32 F(c1) + F(c2) = (9/5 * c1 + 32) + (9/5 * c2 + 32) = 9/5 * (c1 + c2) + 64 F(k*c1) = 9/5 * (k*c1) + 32 k*F(c1) = k * (9/5 * c1 + 32) = 9/5 * (k*c1) + k*32 Admittedly, I'm not entirely sure why being a non-linear transform gets the OP so riled up. [1]http://mathworld.wolfram.com/LinearTransformation.html http://mathworld.wolfram.com/LinearTransformation.html

raverbashing 12y ago
" Admittedly, I'm not entirely sure why being a non-linear transform gets the OP so riled up." Because it's harder to calculate mentally One can approximate 2lb to 1kg roughly, 1mi ~ 1.6km, but for Fahrenheit this is harder One solution is to establish a set point ( 70F = 21.1C, approximate to 20C ) then work in differences, since 1.8deltaF = 1deltaC so, 2dF ~ 1dC (or 10dF ~ 5dC) Hence, 90F is approximately 30C (exact value, 32.2C)